Modulus of Two Complex Numbers, Squared
I have a very silly question to ask! I have
$|z_{1} + z_{2}|^2 = |z_{1}|^2+|z_{2}|^2+2|z_{1}||z_{2}|\cos{\theta}$,
where $z_{1}$ and $z_{2}$ are complex numbers.
For the life of me I cannot reproduce the $\cos{\theta}$ term, to me this looks like the geometric definition of the dot-product but I am unsure as to whether you can treat a complex number in the same way as a "traditional" vector.
I also tried to use Euler's identity to see if the $\cos{\theta}$ term could be produced via its exponential form - no luck!
Any thoughts are appreciated!
$\endgroup$ 21 Answer
$\begingroup$Use conjugates:$$|z_1+z_2|^2 =(z_1+z_2)(\overline{z_1}+\overline{z_2}) =z_1\overline{z_1}+z_2\overline{z_2}+z_1\overline{z_2}+\overline{z_1}z_2\ ,$$that is,$$|z_1+z_2|^2 =|z_1|^2+|z_2|^2+2\Re(z_1\overline{z_2})\ .$$Now if $z_1=re^{i\alpha}$ and $z_2=se^{i\beta}$ then$$\Re(z_1\overline{z_2})=\Re\{rse^{i(\alpha-\beta)}\}=rs\cos(\alpha-\beta)=|z_1||z_2|\cos\theta\ .$$
$\endgroup$ 3
